Find the area of the region, R, bounded by the curve y=x^(-2/3), the line x = 1 and the x axis . In addition, find the volume of revolution of this region when rotated 2 pi radians around the x axis.

As we are looking to find the area under a curve we can use integration

The area under a curve with equation y = f(x), bounded by the lines x = a and x = b and the x axis can be expressed as:

A = integral (from a to b) f(x) dx

Thus, the area of the region, R, can be expressed as:

R = integral (from 1 to infinity) x^(-2/3) dx

R = [3*x^(1/3)] (from 1 to infinity)

R = (3*infinity(1/3))-(3*1^(1/3)) = infinity - 3 = infinity

R = infinity

Therefore the region has infinite area, 

Considering now the volume of revolution, again using integration:

The volume of revolution 2pi radians around the x axis of the same region described above can be expressed:

V = pi * integral (from a to b) f(x)^2 dx

Thus, for the curve in question:

V = pi * integral (from 1 to infinity) [x^(-2/3)]^2 dx = pi * integral (from 1 to infinity) x^(4/3) dx

V = [-3*x^(-1/3)] (from 1 to infinity) = (-3*infinity^(-1/3))-(-3*1^(-1/3)) = 0 - (-3) = 3

V = 3

This interesting problem shows that a region can have infinity area but its revolution can have fininte volume 

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